Answer:
0.792g of triphenyl bromide are produced.
Explanation:
The reaction of triphenyl methanol with HBr is:
triphenyl methanol + HBr → Triphenyl bromide.
<em>Reaction (1:1), 1 mole of HBr reacts per mole of triphenyl methanol.</em>
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To know the mass of triphenyl bromide assuming a theoretical yield (Yield 100%) we need to find first <em>limiting reactant</em>:
Moles triphenyl methanol (Molar mass: 260.33g/mol) =
0.220g × (1mol / 260.33g) = <em>8.45x10⁻³ moles Triphenyl methanol</em>
Moles HBr (Molar mass: 80.91g/mol; 33%=33g HBr/100mL) =
0.60mL ₓ (33g / 100mL) ₓ (1mol / 80.91g) = <em>2.45x10⁻³ moles HBr</em>
<em />
As amount of moles of HBr is lower than moles of triphenyl methanol, HBr is <em>limiting reactant.</em>
<em />
As HBr is limiting reactant, moles produced of triphenyl bromide = moles HBr = <em>2.45x10⁻³ moles</em>
As molar mass of triphenyl bromide is 323.2g/mol, mass of triphenyl bromide is:
2.45x10⁻³ moles × (323.2g / mol) =
<h3>0.792g of triphenyl bromide are produced.</h3>
